The total number of electrons in d orbital in Fe2+ ion (atomic number of Fe is 26) is not equal to that of the total number of :

The number that differs is the number of p electrons in the Cl− ion.Fe is [Ar] 3d6 4s2; Fe2+ loses the two 4s electrons and keeps 3d6, so it has 6 d electrons.Ne, 1s2 2s2 2p6: 6 p electrons.Fe atom, 3d6: 6 d electrons.Mg, 1s2 2s2 2p6 3s2: 6 s electrons.Cl−, 18 electrons, 1s2 2s2 2p6 3s2 3p6: 12 p electrons.Only Cl− does not match the 6 of Fe2+.

Many-electron Atoms and Coupling Schemes Questions and Answers

Q: Which one of the following is an allowed electric dipole transition?

The allowed transition is \({}^{2}\mathrm{P}_{3/2} \rightarrow {}^{2}\mathrm{D}_{5/2}\).Electric dipole (LS coupling) selection rules: \(\Delta S = 0\), \(\Delta L = 0, \pm 1\) (with a one-electron jump changing \(l\) by \(\pm 1\)), \(\Delta J = 0, \pm 1\), and \(J = 0 \to 0\) forbidden.For \({}^{2}\mathrm{P}_{3/2} \to {}^{2}\mathrm{D}_{5/2}\): \(\Delta S = 0\), \(\Delta L = +1\), \(\Delta J = +1\). All satisfied.The others each break a rule. \({}^{1}\mathrm{S}_{0} \to {}^{3}\mathrm{S}_{1}\) changes the spin multiplicity and has \(\Delta L = 0\) between S states. \({}^{2}\mathrm{D}_{5/2} \to {}^{2}\mathrm{P}_{1/2}\) has \(\Delta J = 2\). \({}^{3}\mathrm{P}_{0} \to {}^{5}\mathrm{D}_{0}\) changes the spin and is a \(J = 0 \to 0\) transition.

Q: The term symbol for the electronic ground state of oxygen atom is

The ground-state term of oxygen is \({}^{3}P_{2}\).Oxygen is \(1s^{2}2s^{2}2p^{4}\). Hund’s rules for the four 2p electrons:Maximum spin: place the electrons with \(m_{l} = +1, 0, -1\) spin up and one more spin down, giving \(S = 1\), a triplet.Maximum \(L\) for that spin: the paired electron goes into \(m_{l} = +1\), so \(M_{L} = 1\) and \(L = 1\), a P term.The shell is more than half full, so the largest \(J = L + S = 2\) lies lowest.\({}^{3}P_{0}\) would be the ground state for a less-than-half-filled p shell such as carbon (\(2p^{2}\)). \({}^{1}D_{2}\) and \({}^{1}S_{0}\) are excited terms of the same configuration, with lower spin.

Q: An excited state of Ca atom is [Mg]3p54s23d1. The spectroscopic terms corresponding to the total orbital angular momentum are

The terms are P, D, and F.Closed shells and the filled 4s2 subshell contribute no angular momentum. What remains is a 3p5 configuration, which behaves like a single p hole with \(l_{1} = 1\), and one 3d electron with \(l_{2} = 2\). Their orbital angular momenta couple to\[ L = |l_{1} - l_{2}|, \ldots, l_{1} + l_{2} = 1, 2, 3 \]giving P, D and F terms. S would need \(L = 0\), impossible when \(l_{1} \ne l_{2}\), so the options containing S are wrong, and 'P and D' omits \(L = 3\).

Q: The atomic number of an atom is 6. What is the spectroscopic notation of its ground state, according to Hund’s rules?

The ground state is \({}^{3}P_{0}\).Atomic number 6 is carbon, \(1s^{2}2s^{2}2p^{2}\). Only the two 2p electrons matter. Hund's rules then fix the term:Maximum spin: the electrons take parallel spins, so \(S = 1\) and the multiplicity is \(2S + 1 = 3\).Maximum orbital angular momentum allowed with parallel spins: \(m_{l} = +1\) and \(0\), so \(L = 1\), a P term.The subshell is less than half filled, so the lowest level has the smallest \(J = |L - S| = 0\).\({}^{3}P_{1}\) would apply only if \(J\) took the middle value, and \({}^{3}P_{2}\) would be the ground state for a more than half-filled shell such as oxygen. \({}^{3}D_{3}\) needs \(L = 2\), impossible for two parallel p electrons, and \({}^{3}S_{1}\) needs \(L = 0\).

Q: The possible values of j and mj for a single d electron system would be :

The values are \( j = \frac{5}{2} \) and \( \frac{3}{2} \), with \( m_{j} = \pm\frac{5}{2}, \pm\frac{3}{2}, \pm\frac{1}{2} \).A d electron has \( l = 2 \) and \( s = \frac{1}{2} \), so \( j = l \pm s = \frac{5}{2} \) or \( \frac{3}{2} \). For each \( j \), \( m_{j} \) runs from \( -j \) to \( +j \) in steps of 1, so together the values are \( \pm\frac{5}{2}, \pm\frac{3}{2}, \pm\frac{1}{2} \).Integer \( j \) values such as 2, 1 or 3 are impossible for a single electron, whose half-integer spin makes \( j \) half-integral. \( m_{j} = 5, 3, 1, \ldots \) doubles the correct values, and \( m_{j} \) must lie between \( -j \) and \( +j \).

Q: The S, L and J values that correspond to each of the following states \(^{1}S_{0}\), \(^{3}P_{2}\) are :

The values are S = 0, L = 0, J = 0 and S = 1, L = 1, J = 2.A term symbol \(^{2S+1}L_{J}\) gives the multiplicity 2S + 1 as a superscript, L as a letter (S = 0, P = 1, D = 2) and J as a subscript.For \(^{1}S_{0}\): 2S + 1 = 1 so S = 0, the letter S means L = 0, and J = 0. For \(^{3}P_{2}\): 2S + 1 = 3 so S = 1, P means L = 1, and J = 2. Option B reads the superscript as S itself, giving S = 1 and S = 2, and options C and D mix up L.

Q: Which of the both sequences represent an increasing order of orbital energy ?

The sequences in increasing order are 3p, 4s, 3d, 4p & 2p, 3s, 3p, 4s.By the (n + l) rule, orbitals fill in order of increasing n + l, and for equal n + l the one with lower n comes first: 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d and so on. Both sequences in option D follow this order.Option A's second sequence puts 3d before 4s. Option B puts 3d before 4s in its first sequence. Option C's second sequence puts 5s before 3d and 4f before 4p, both reversed.

Q: Te2+ ion has two d electrons in its outermost shell. With the aid of Hund's rule the ground term for the d2 system will be :

The ground term is \(^{3}F\), with \(M_{L} = 3\) and \(M_{S} = 1\).Apply Hund's rules to two d electrons (\(m_{l} = 2, 1, 0, -1, -2\)):Maximise S: put the two electrons in different orbitals with parallel spins, so \(M_{S} = \frac{1}{2} + \frac{1}{2} = 1\), S = 1, multiplicity 3.Then maximise L: place them in \(m_{l} = 2\) and \(m_{l} = 1\), so \(M_{L} = 3\), L = 3, an F term.The ground term is \(^{3}F\) (with J = 2 for a shell less than half full, \(^{3}F_{2}\)), as for \(Ti^{2+}\) or \(V^{3+}\). A doublet such as \(^{2}D\) belongs to a single d electron, and a quintet (\(M_{S} = 2\)) would need four unpaired electrons.

Which one of the following is an allowed electric dipole transition?

A.\({}^{1}\mathrm{S}_{0} \rightarrow {}^{3}\mathrm{S}_{1}\)
B.\({}^{2}\mathrm{P}_{3/2} \rightarrow {}^{2}\mathrm{D}_{5/2}\)
C.\({}^{2}\mathrm{D}_{5/2} \rightarrow {}^{2}\mathrm{P}_{1/2}\)
D.\({}^{3}\mathrm{P}_{0} \rightarrow {}^{5}\mathrm{D}_{0}\)

Exam Year(s)

GATE Physics - 2018,

Answer: B

The allowed transition is \({}^{2}\mathrm{P}_{3/2} \rightarrow {}^{2}\mathrm{D}_{5/2}\).

Electric dipole (LS coupling) selection rules: \(\Delta S = 0\), \(\Delta L = 0, \pm 1\) (with a one-electron jump changing \(l\) by \(\pm 1\)), \(\Delta J = 0, \pm 1\), and \(J = 0 \to 0\) forbidden.

For \({}^{2}\mathrm{P}_{3/2} \to {}^{2}\mathrm{D}_{5/2}\): \(\Delta S = 0\), \(\Delta L = +1\), \(\Delta J = +1\). All satisfied.

The others each break a rule. \({}^{1}\mathrm{S}_{0} \to {}^{3}\mathrm{S}_{1}\) changes the spin multiplicity and has \(\Delta L = 0\) between S states. \({}^{2}\mathrm{D}_{5/2} \to {}^{2}\mathrm{P}_{1/2}\) has \(\Delta J = 2\). \({}^{3}\mathrm{P}_{0} \to {}^{5}\mathrm{D}_{0}\) changes the spin and is a \(J = 0 \to 0\) transition.

Discuss in Forum

The term symbol for the electronic ground state of oxygen atom is

A.\({}^{1}S_{0}\)
B.\({}^{1}D_{2}\)
C.\({}^{3}P_{0}\)
D.\({}^{3}P_{2}\)

Exam Year(s)

GATE Physics - 2018,

Answer: D

The ground-state term of oxygen is \({}^{3}P_{2}\).

Oxygen is \(1s^{2}2s^{2}2p^{4}\). Hund’s rules for the four 2p electrons:

Maximum spin: place the electrons with \(m_{l} = +1, 0, -1\) spin up and one more spin down, giving \(S = 1\), a triplet.
Maximum \(L\) for that spin: the paired electron goes into \(m_{l} = +1\), so \(M_{L} = 1\) and \(L = 1\), a P term.
The shell is more than half full, so the largest \(J = L + S = 2\) lies lowest.

\({}^{3}P_{0}\) would be the ground state for a less-than-half-filled p shell such as carbon (\(2p^{2}\)). \({}^{1}D_{2}\) and \({}^{1}S_{0}\) are excited terms of the same configuration, with lower spin.

Discuss in Forum

An excited state of Ca atom is [Mg]3p⁵4s²3d¹. The spectroscopic terms corresponding to the total orbital angular momentum are

A.S, P, and D
B.P, D, and F
C.P and D
D.S and P

Exam Year(s)

GATE Physics - 2022,

Answer: B

The terms are P, D, and F.

Closed shells and the filled 4s² subshell contribute no angular momentum. What remains is a 3p⁵ configuration, which behaves like a single p hole with \(l_{1} = 1\), and one 3d electron with \(l_{2} = 2\). Their orbital angular momenta couple to

\[ L = |l_{1} - l_{2}|, \ldots, l_{1} + l_{2} = 1, 2, 3 \]

giving P, D and F terms. S would need \(L = 0\), impossible when \(l_{1} \ne l_{2}\), so the options containing S are wrong, and 'P and D' omits \(L = 3\).

Discuss in Forum

The atomic number of an atom is 6. What is the spectroscopic notation of its ground state, according to Hund’s rules?

A.\({}^{3}P_{0}\)
B.\({}^{3}P_{1}\)
C.\({}^{3}D_{3}\)
D.\({}^{3}S_{1}\)

Exam Year(s)

GATE Physics - 2023,

Answer: A

The ground state is \({}^{3}P_{0}\).

Atomic number 6 is carbon, \(1s^{2}2s^{2}2p^{2}\). Only the two 2p electrons matter. Hund's rules then fix the term:

Maximum spin: the electrons take parallel spins, so \(S = 1\) and the multiplicity is \(2S + 1 = 3\).
Maximum orbital angular momentum allowed with parallel spins: \(m_{l} = +1\) and \(0\), so \(L = 1\), a P term.
The subshell is less than half filled, so the lowest level has the smallest \(J = |L - S| = 0\).

\({}^{3}P_{1}\) would apply only if \(J\) took the middle value, and \({}^{3}P_{2}\) would be the ground state for a more than half-filled shell such as oxygen. \({}^{3}D_{3}\) needs \(L = 2\), impossible for two parallel p electrons, and \({}^{3}S_{1}\) needs \(L = 0\).

Discuss in Forum

The X-rays emitted by a target consist of continuous range of wavelengths superimposed by characteristic X-rays named \(K_{\alpha}\), \(K_{\beta}\) etc. It is found that :

A.\(K_{\alpha}\) has high intensity and lower wavelength compared to that of \(K_{\beta}\).
B.\(K_{\alpha}\) has low intensity and lower wavelength as compared to that of \(K_{\beta}\).
C.\(K_{\alpha}\) has high intensity and higher wavelength compared to that of \(K_{\beta}\).
D.\(K_{\alpha}\) has lower intensity and higher wavelength compared to that of \(K_{\beta}\).

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

\(K_{\alpha}\) has high intensity and higher wavelength compared to \(K_{\beta}\).

\(K_{\alpha}\) lines come from electrons falling from the L shell into a K-shell vacancy, and \(K_{\beta}\) from the M shell. The L to K gap is smaller than the M to K gap, so the \(K_{\alpha}\) photon has less energy and therefore a longer wavelength, since \(\lambda = hc/E\).

The L to K transition is also much more probable than M to K, because the L shell is closer and more strongly coupled, so \(K_{\alpha}\) is the more intense line, typically several times stronger than \(K_{\beta}\). The other options reverse one or both of these comparisons.

Discuss in Forum

The total number of electrons in d orbital in Fe²⁺ ion (atomic number of Fe is 26) is not equal to that of the total number of :

A.p electrons in Ne atom (Atomic number 10)
B.d electrons in Fe atom
C.p electrons in \( \text{Cl}^{-} \) ion (Atomic number 17)
D.s electrons of Mg (Atomic number 12)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The number that differs is the number of p electrons in the Cl⁻ ion.

Fe is [Ar] 3d⁶ 4s²; Fe²⁺ loses the two 4s electrons and keeps 3d⁶, so it has 6 d electrons.

Ne, 1s² 2s² 2p⁶: 6 p electrons.
Fe atom, 3d⁶: 6 d electrons.
Mg, 1s² 2s² 2p⁶ 3s²: 6 s electrons.
Cl⁻, 18 electrons, 1s² 2s² 2p⁶ 3s² 3p⁶: 12 p electrons.

Only Cl⁻ does not match the 6 of Fe²⁺.

Discuss in Forum

The possible values of j and m_j for a single d electron system would be :

A.\( j = 2, 1 \) and \( m_{j} = \frac{5}{2} \) and \( \frac{3}{2} \)
B.\( j = \frac{5}{2} \) and \( \frac{3}{2} \) and \( m_{j} = \frac{5}{2}, \frac{3}{2}, \frac{1}{2}, \) \( \frac{-1}{2}, \frac{-3}{2}, \frac{-5}{2} \)
C.\( j = 3 \) and 2 and \( m_{j} = 3, 2, 1, 0, -1, -2, -3 \)
D.\( j = \frac{5}{2} \) and \( \frac{3}{2} \) and \( m_{j} = 5, 3, 1, -1, -3, -5 \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The values are \( j = \frac{5}{2} \) and \( \frac{3}{2} \), with \( m_{j} = \pm\frac{5}{2}, \pm\frac{3}{2}, \pm\frac{1}{2} \).

A d electron has \( l = 2 \) and \( s = \frac{1}{2} \), so \( j = l \pm s = \frac{5}{2} \) or \( \frac{3}{2} \). For each \( j \), \( m_{j} \) runs from \( -j \) to \( +j \) in steps of 1, so together the values are \( \pm\frac{5}{2}, \pm\frac{3}{2}, \pm\frac{1}{2} \).

Integer \( j \) values such as 2, 1 or 3 are impossible for a single electron, whose half-integer spin makes \( j \) half-integral. \( m_{j} = 5, 3, 1, \ldots \) doubles the correct values, and \( m_{j} \) must lie between \( -j \) and \( +j \).

Discuss in Forum

The S, L and J values that correspond to each of the following states \(^{1}S_{0}\), \(^{3}P_{2}\) are :

A.S = 0, L = 0, J = 0 and S = 1, L = 1, J = 2
B.S = 1, L = 0, J = 0 and S = 2, L = 1, J = 2
C.S = 0, L = 1, J = 1 and S = 1, L = 0, J = 2
D.S = 1, L = 1, J = 2 and S = 2, L = 0, J = 2

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The values are S = 0, L = 0, J = 0 and S = 1, L = 1, J = 2.

A term symbol \(^{2S+1}L_{J}\) gives the multiplicity 2S + 1 as a superscript, L as a letter (S = 0, P = 1, D = 2) and J as a subscript.

For \(^{1}S_{0}\): 2S + 1 = 1 so S = 0, the letter S means L = 0, and J = 0. For \(^{3}P_{2}\): 2S + 1 = 3 so S = 1, P means L = 1, and J = 2. Option B reads the superscript as S itself, giving S = 1 and S = 2, and options C and D mix up L.

Discuss in Forum

Which of the both sequences represent an increasing order of orbital energy ?

A.2p, 3s, 3p, 3d & 3s, 3p, 3d, 4s
B.3s, 3p, 3d, 4s & 4s, 5s, 6s, 7s
C.3s, 3p, 4s, 3d & 5s, 3d, 4f, 4p
D.3p, 4s, 3d, 4p & 2p, 3s, 3p, 4s

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The sequences in increasing order are 3p, 4s, 3d, 4p & 2p, 3s, 3p, 4s.

By the (n + l) rule, orbitals fill in order of increasing n + l, and for equal n + l the one with lower n comes first: 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d and so on. Both sequences in option D follow this order.

Option A's second sequence puts 3d before 4s. Option B puts 3d before 4s in its first sequence. Option C's second sequence puts 5s before 3d and 4f before 4p, both reversed.

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10.

T_e²⁺ ion has two d electrons in its outermost shell. With the aid of Hund's rule the ground term for the d² system will be :

A.\(M_{L} = 2,\ M_{S} = 1/2\), Term = \(^{2}D\)
B.\(M_{L} = 3,\ M_{S} = 1\), Term = \(^{3}F\)
C.\(M_{L} = 2,\ M_{S} = 2\), Term = \(^{5}D\)
D.\(M_{L} = 2,\ M_{S} = -1/2\), Term = \(^{2}D\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The ground term is \(^{3}F\), with \(M_{L} = 3\) and \(M_{S} = 1\).

Apply Hund's rules to two d electrons (\(m_{l} = 2, 1, 0, -1, -2\)):

Maximise S: put the two electrons in different orbitals with parallel spins, so \(M_{S} = \frac{1}{2} + \frac{1}{2} = 1\), S = 1, multiplicity 3.
Then maximise L: place them in \(m_{l} = 2\) and \(m_{l} = 1\), so \(M_{L} = 3\), L = 3, an F term.

The ground term is \(^{3}F\) (with J = 2 for a shell less than half full, \(^{3}F_{2}\)), as for \(Ti^{2+}\) or \(V^{3+}\). A doublet such as \(^{2}D\) belongs to a single d electron, and a quintet (\(M_{S} = 2\)) would need four unpaired electrons.

Many-electron Atoms and Coupling Schemes Questions and Answers

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